# Complex equation

• Jun 20th 2010, 12:29 AM
Mondreus
Complex equation
Solve: $e^{z^2}=i$

I tried writing it like this: $e^{z^2} = e^{e^{2\log(z)}}=e^{\pi i /2+2\pi i n$

That leaves us with: $e^{2\log(z)} = \pi i /2+2\pi i n$, but that didn't help much.
• Jun 20th 2010, 01:15 AM
Defunkt
use the identity $i = e^{\frac{i \pi}{2} + 2 \pi i k}, k = 0, \pm 1, \pm 2 , \ldots$
• Jun 20th 2010, 01:21 AM
Mondreus
I've already used it, and it leads me to the last equation.
• Jun 20th 2010, 01:58 AM
Prove It
Quote:

Originally Posted by Mondreus
Solve: $e^{z^2}=i$

I tried writing it like this: $e^{z^2} = e^{e^{2\log(z)}}=e^{\pi i /2+2\pi i n$

That leaves us with: $e^{2\log(z)} = \pi i /2+2\pi i n$, but that didn't help much.

$e^{z^2} = i$

$e^{(x + iy)^2} = i$

$e^{x^2 + 2ixy +i^2y^2} = i$

$e^{x^2 - y^2 + 2ixy} = i$

$e^{x^2 - y^2}\,\textrm{cis}\,(2xy) = i$

$e^{x^2 - y^2}\cos{(2xy)} + i\,e^{x^2 - y^2}\sin{(2xy)} = 0 + i$.

Equating real and imaginary parts gives:

$e^{x^2 - y^2}\cos{(2xy)} = 0$ and $e^{x^2 - y^2}\sin{(2xy)} = 1$.

Equation 1:

$e^{x^2 - y^2}\cos{(2xy)} = 0$

Since the exponential function is always positive, this means

$\cos{(2xy)} = 0$.

From the Pythagorean Identity:

$\cos^2{(2xy)} + \sin^2{(2xy)} = 1$

$\sin^2{(2xy)} = 1$ since $\cos{(2xy)} = 0$.

Can you go from here?
• Jun 20th 2010, 03:07 AM
Mondreus
The answer is $z=\pm(1+i)\sqrt{\pi/4+n\pi}$ and $z=\pm(1-i)\sqrt{-\pi/4+n\pi}$, but I can't there from what you posted.
• Jun 20th 2010, 07:34 PM
Prove It
Quote:

Originally Posted by Prove It
$e^{z^2} = i$

$e^{(x + iy)^2} = i$

$e^{x^2 + 2ixy +i^2y^2} = i$

$e^{x^2 - y^2 + 2ixy} = i$

$e^{x^2 - y^2}\,\textrm{cis}\,(2xy) = i$

$e^{x^2 - y^2}\cos{(2xy)} + i\,e^{x^2 - y^2}\sin{(2xy)} = 0 + i$.

Equating real and imaginary parts gives:

$e^{x^2 - y^2}\cos{(2xy)} = 0$ and $e^{x^2 - y^2}\sin{(2xy)} = 1$.

Equation 1:

$e^{x^2 - y^2}\cos{(2xy)} = 0$

Since the exponential function is always positive, this means

$\cos{(2xy)} = 0$.

From the Pythagorean Identity:

$\cos^2{(2xy)} + \sin^2{(2xy)} = 1$

$\sin^2{(2xy)} = 1$ since $\cos{(2xy)} = 0$.

Can you go from here?

Well, from where I left off...

$\sin^2{(2xy)} = 1$ since $\cos{(2xy)} = 0$

$\sin{(2xy)} = 1$ or $\sin{(2xy)} = -1$.

Substituting into equation 2:

$e^{x^2 - y^2}\sin{(2xy)} = 1$

Since $e^{x^2 - y^2}$ is always positive, that means the only acceptable solution is $\sin{(2xy)} = 1$.

So $e^{x^2 - y^2} = 1$

$x^2 - y^2 = \ln{1}$

$x^2 - y^2 = 0$

$y^2 = x^2$

$y = \pm x$.

Subtituting back into $\cos{(2xy)} = 0$

$\cos{[2x(\pm x)]} = 0$

$\cos{(\pm 2x^2)} = 0$

$\pm 2x^2 = \frac{\pi}{2} + \pi n$ where $n \in \mathbf{Z}$

$x^2 = \pm\left(\frac{\pi}{4} + \frac{\pi n}{2}\right)$

$x = \pm \sqrt{ \pm(\frac{\pi}{4} + \frac{\pi n}{2} \right ) }$.

Since $y = \pm$ that means $y = \pm \sqrt{\pm \left(\frac{\pi}{4} + \frac{\pi n}{2}\right)}$.

Can you go from here?
• Jun 20th 2010, 10:22 PM
Mondreus
That seems to be more solutions than what the textbook says though. This is how I "solved" it, but I'm not sure everything I've done is correct, and I seem to be missing some solutions. I'm especially confused about taking the square root of a complex number, and when you can factor out complex numbers from square roots.

$z^2 = i(\pi /2 + 2 \pi n) \Leftrightarrow z^2 = 2e^{i(\pi /2 + 2 \pi k)}(\pi / 4 + \pi n) \Leftrightarrow$

$z = \pm \sqrt{2}\sqrt{e^{i(\pi /2 + 2 \pi k)}(\pi / 4 + \pi n)} = \pm \sqrt{2}\sqrt{e^{i(\pi /2 + 2 \pi k)}}\sqrt{\pi / 4 + \pi n} = \pm(1+i)\sqrt{\pi / 4 + \pi n}$

I know that this isn't true in general if z and w are complex numbers: $\sqrt{z}\sqrt{w} = \sqrt{zw}$. It's often useful to factor out $-1 = i^2$ from square roots though, so how do you know when you can do it?